Leetcode #2423: Remove Letter To Equalize Frequency
In this guide, we solve Leetcode #2423 Remove Letter To Equalize Frequency in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Hash Table, String, Counting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: word = "abcc"
Output: true
Explanation: Select index 3 and delete it: word becomes "abc" and each character has a frequency of 1.
Python Solution
class Solution:
def equalFrequency(self, word: str) -> bool:
cnt = Counter(word)
for c in cnt.keys():
cnt[c] -= 1
if len(set(v for v in cnt.values() if v)) == 1:
return True
cnt[c] += 1
return False
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.